Consider two square, equilateral pyramids of equal volume $V_{object}$ and equal mass.

  1. One pyramid is uniformly dense
  2. The other pyramid has most of its mass concentrated at the apex.

Both pyramids have an (equal) overall density $\rho_{object}$ less than that of water ($\rho_{fluid}$), so they float. Will the pyramids float with a different amount submerged, and if so, why?

On one hand, I know that objects float when the gravitational force on the object is equal and opposite to the buoyant force of the water on the object. This implies that the mass of the object and the displaced water are equal, or

$$ \rho_{fluid} V_{displaced} = \rho_{object} V_{object}. $$ Since my rearranging the location of the mass in the object without changing its density, we know that $\rho_{fluid}$, $\rho_{object}$, and $V_{object}$ are all held constant, so the volume displaced must remain the same.

On the other hand perhaps the fact that the denser part of the pyramid will likely face down when submerged means that the PART of the pyramid submerged in the water is denser, so we need to increase $\rho_{object}$ in this equation. That would imply that the $V_{displaced}$ would have to increase to compensate.

Is one of these arguments correct, or are they both wrong?


1 Answer 1


Your reasoning in "on one hand" is correct, whereas the other hand is wrong.

That is, the volume of the submerged part of the pyramid will stay the same regardless of how the mass in the pyramid is distributed.

This is like laying down in a canoe vs standing up: the amount of water that the canoe displaces will remain the same.


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